How will you convert ethanoic acid into benzene?

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Solution

I) $C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$
(etharoic acid)
II) $C H_{3} C O O N a N a O H - --- \to C a O; Δ C H_{4} + N a_{2} C O_{3}$
III) $C H_{4} + B H_{2} Δ \to C H_{3} B r + H B r$
IV) $2 C H_{3} B r N a - -- \to c t h e r C H_{3} - C H_{3}$
V) $C_{2} H_{6} + B r_{2} Δ \to C H_{3} - C H_{2} B r$
VI) $C H_{3} - C H_{2} B r a l c . k O H \to C H_{2} = C H_{2}$
VII) $C H_{2} = C H_{2} + B r_{2} c c l_{4} \to$
VIII) $B r C H_{2} - C H_{2} - B r a l c . k O H \to H C \equiv C H$
IX) $3 C H \equiv C H r e d h s t - --- \to F e t u b e$

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